the basic function is f(x) = 1/x

the transformed function that I don't know how to graph is: y = f[(-1/3)x + 1]

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- Apr 27th 2010, 02:30 PMneedhelpplease1How to graph this...
the basic function is f(x) = 1/x

the transformed function that I don't know how to graph is: y = f[(-1/3)x + 1] - Apr 27th 2010, 03:10 PMe^(i*pi)
The way you've written it implies $\displaystyle f[-\frac{x}{3} +1]$ which doesn't fit with the basic function.

I will assume you mean $\displaystyle f[-\frac{1}{3x}+1]$. It's best to do these in a series of steps.

For f(-x) reflect f(x) in the y-axis

For f(ax) increase 'steepness' by a. In this case it means it will come down from $\displaystyle \infty$ more sharply. You're not expected to know the exact shape but you should evaluate f(1) to get a set point. In your case the point will be $\displaystyle \left(1, \frac{4}{3}\right)$

For f(x)+b you move the graph up the y axis by b. In your case b=1. This will also move any asymptotes.

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In your case you want the graph of 1/x:

- Reflected in the y-axis
- Steepened by a factor of 3 - passing through $\displaystyle \left(1, \frac{4}{3}\right)$
- Moved up by 1 unit on the y-axis

See the graph attached for the final answer, I have included 1/x for comparison. - Apr 27th 2010, 03:26 PMneedhelpplease1Thank you thank you!
Thanks you SO much! That was amazingly helpful!!